Orthogonality of the DFT Sinusoids

We now show mathematically that the DFT sinusoids are exactly orthogonal. Let

$$s_k(n) \isdef e^{j\omega_k nT} = e^{j2\pi k n /N} = \left[W_N^k\right]^n$$

denote the $k$th complex DFT sinusoid. Then

$$\begin{eqnarray*} \left<s_k,s_l\right> &\isdef & \sum_{n=0}^{N-1}s_k(n) \overlin... ...i (k-l) n /N} = \frac{1 - e^{j2\pi (k-l)}}{1-e^{j2\pi (k-l)/N}} \end{eqnarray*}$$

where the last step made use of the closed-form expression for the sum of a geometric series. If $k\neq l$, the denominator is nonzero while the numerator is zero. This proves

$$\zbox {s_k \perp s_l, \quad k \neq l}.$$

While we only looked at unit amplitude, zero phase complex sinusoids, as used by the DFT, it is readily verified that the (nonzero) amplitude and phase have no effect on orthogonality.